Problem: Let $x,$ $y,$ $z$ be real numbers such that
\begin{align*}
x + y + z &= 4, \\
x^2 + y^2 + z^2 &= 6.
\end{align*}Let $m$ and $M$ be the smallest and largest possible values of $x,$ respectively.  Find $m + M.$
Solution: From the given equations, $y + z = 4 - x$ and $y^2 + z^2 = 6 - x^2.$  By Cauchy-Schwarz,
\[(1 + 1)(y^2 + z^2) \ge (y + z)^2.\]Hence, $2(6 - x^2) \ge (4 - x)^2.$  This simplifies to $3x^2 - 8x + 4 \le 0,$ which factors as $(x - 2)(3x - 2) \le 0.$  Hence, $\frac{2}{3} \le x \le 2.$

For $x = \frac{3}{2},$ we can take $y = z = \frac{5}{3}.$  For $x = 2,$ we can take $y = z = 1.$  Thus, $m = \frac{2}{3}$ and $M = 2,$ so $m + M = \boxed{\frac{8}{3}}.$